### 3.323 $$\int \frac{(d+e x)^2}{(b x+c x^2)^{3/2}} \, dx$$

Optimal. Leaf size=101 $\frac{2 e \sqrt{b x+c x^2} (2 c d-b e)}{b^2 c}-\frac{2 (d+e x) (x (2 c d-b e)+b d)}{b^2 \sqrt{b x+c x^2}}+\frac{2 e^2 \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{c^{3/2}}$

[Out]

(-2*(d + e*x)*(b*d + (2*c*d - b*e)*x))/(b^2*Sqrt[b*x + c*x^2]) + (2*e*(2*c*d - b*e)*Sqrt[b*x + c*x^2])/(b^2*c)
+ (2*e^2*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/c^(3/2)

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Rubi [A]  time = 0.061544, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.19, Rules used = {738, 640, 620, 206} $\frac{2 e \sqrt{b x+c x^2} (2 c d-b e)}{b^2 c}-\frac{2 (d+e x) (x (2 c d-b e)+b d)}{b^2 \sqrt{b x+c x^2}}+\frac{2 e^2 \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{c^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[(d + e*x)^2/(b*x + c*x^2)^(3/2),x]

[Out]

(-2*(d + e*x)*(b*d + (2*c*d - b*e)*x))/(b^2*Sqrt[b*x + c*x^2]) + (2*e*(2*c*d - b*e)*Sqrt[b*x + c*x^2])/(b^2*c)
+ (2*e^2*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/c^(3/2)

Rule 738

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m - 1)*(
d*b - 2*a*e + (2*c*d - b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)), x] + Dist[1/((p + 1)*(b^2 -
4*a*c)), Int[(d + e*x)^(m - 2)*Simp[e*(2*a*e*(m - 1) + b*d*(2*p - m + 4)) - 2*c*d^2*(2*p + 3) + e*(b*e - 2*d*
c)*(m + 2*p + 2)*x, x]*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] &
& NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && LtQ[p, -1] && GtQ[m, 1] && IntQuadraticQ[a, b, c, d,
e, m, p, x]

Rule 640

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(a + b*x + c*x^2)^(p +
1))/(2*c*(p + 1)), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rule 620

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2
]], x] /; FreeQ[{b, c}, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{(d+e x)^2}{\left (b x+c x^2\right )^{3/2}} \, dx &=-\frac{2 (d+e x) (b d+(2 c d-b e) x)}{b^2 \sqrt{b x+c x^2}}-\frac{2 \int \frac{-b d e-e (2 c d-b e) x}{\sqrt{b x+c x^2}} \, dx}{b^2}\\ &=-\frac{2 (d+e x) (b d+(2 c d-b e) x)}{b^2 \sqrt{b x+c x^2}}+\frac{2 e (2 c d-b e) \sqrt{b x+c x^2}}{b^2 c}+\frac{e^2 \int \frac{1}{\sqrt{b x+c x^2}} \, dx}{c}\\ &=-\frac{2 (d+e x) (b d+(2 c d-b e) x)}{b^2 \sqrt{b x+c x^2}}+\frac{2 e (2 c d-b e) \sqrt{b x+c x^2}}{b^2 c}+\frac{\left (2 e^2\right ) \operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,\frac{x}{\sqrt{b x+c x^2}}\right )}{c}\\ &=-\frac{2 (d+e x) (b d+(2 c d-b e) x)}{b^2 \sqrt{b x+c x^2}}+\frac{2 e (2 c d-b e) \sqrt{b x+c x^2}}{b^2 c}+\frac{2 e^2 \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{c^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0960489, size = 100, normalized size = 0.99 $\frac{2 b^{5/2} e^2 \sqrt{x} \sqrt{\frac{c x}{b}+1} \sinh ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )-2 \sqrt{c} \left (b^2 e^2 x+b c d (d-2 e x)+2 c^2 d^2 x\right )}{b^2 c^{3/2} \sqrt{x (b+c x)}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(d + e*x)^2/(b*x + c*x^2)^(3/2),x]

[Out]

(-2*Sqrt[c]*(2*c^2*d^2*x + b^2*e^2*x + b*c*d*(d - 2*e*x)) + 2*b^(5/2)*e^2*Sqrt[x]*Sqrt[1 + (c*x)/b]*ArcSinh[(S
qrt[c]*Sqrt[x])/Sqrt[b]])/(b^2*c^(3/2)*Sqrt[x*(b + c*x)])

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Maple [A]  time = 0.061, size = 97, normalized size = 1. \begin{align*} -2\,{\frac{{e}^{2}x}{c\sqrt{c{x}^{2}+bx}}}+{{e}^{2}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){c}^{-{\frac{3}{2}}}}+4\,{\frac{dex}{b\sqrt{c{x}^{2}+bx}}}-2\,{\frac{{d}^{2} \left ( 2\,cx+b \right ) }{{b}^{2}\sqrt{c{x}^{2}+bx}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^2/(c*x^2+b*x)^(3/2),x)

[Out]

-2*e^2/c/(c*x^2+b*x)^(1/2)*x+e^2/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x)^(1/2))+4*d*e/b/(c*x^2+b*x)^(1/2)*x
-2*d^2*(2*c*x+b)/b^2/(c*x^2+b*x)^(1/2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2/(c*x^2+b*x)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.05542, size = 505, normalized size = 5. \begin{align*} \left [\frac{{\left (b^{2} c e^{2} x^{2} + b^{3} e^{2} x\right )} \sqrt{c} \log \left (2 \, c x + b + 2 \, \sqrt{c x^{2} + b x} \sqrt{c}\right ) - 2 \,{\left (b c^{2} d^{2} +{\left (2 \, c^{3} d^{2} - 2 \, b c^{2} d e + b^{2} c e^{2}\right )} x\right )} \sqrt{c x^{2} + b x}}{b^{2} c^{3} x^{2} + b^{3} c^{2} x}, -\frac{2 \,{\left ({\left (b^{2} c e^{2} x^{2} + b^{3} e^{2} x\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{c x^{2} + b x} \sqrt{-c}}{c x}\right ) +{\left (b c^{2} d^{2} +{\left (2 \, c^{3} d^{2} - 2 \, b c^{2} d e + b^{2} c e^{2}\right )} x\right )} \sqrt{c x^{2} + b x}\right )}}{b^{2} c^{3} x^{2} + b^{3} c^{2} x}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2/(c*x^2+b*x)^(3/2),x, algorithm="fricas")

[Out]

[((b^2*c*e^2*x^2 + b^3*e^2*x)*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 2*(b*c^2*d^2 + (2*c^3*d^2
- 2*b*c^2*d*e + b^2*c*e^2)*x)*sqrt(c*x^2 + b*x))/(b^2*c^3*x^2 + b^3*c^2*x), -2*((b^2*c*e^2*x^2 + b^3*e^2*x)*s
qrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) + (b*c^2*d^2 + (2*c^3*d^2 - 2*b*c^2*d*e + b^2*c*e^2)*x)*sqrt(
c*x^2 + b*x))/(b^2*c^3*x^2 + b^3*c^2*x)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d + e x\right )^{2}}{\left (x \left (b + c x\right )\right )^{\frac{3}{2}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**2/(c*x**2+b*x)**(3/2),x)

[Out]

Integral((d + e*x)**2/(x*(b + c*x))**(3/2), x)

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Giac [A]  time = 1.26464, size = 120, normalized size = 1.19 \begin{align*} -\frac{2 \,{\left (\frac{d^{2}}{b} + \frac{{\left (2 \, c^{2} d^{2} - 2 \, b c d e + b^{2} e^{2}\right )} x}{b^{2} c}\right )}}{\sqrt{c x^{2} + b x}} - \frac{e^{2} \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} \sqrt{c} - b \right |}\right )}{c^{\frac{3}{2}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2/(c*x^2+b*x)^(3/2),x, algorithm="giac")

[Out]

-2*(d^2/b + (2*c^2*d^2 - 2*b*c*d*e + b^2*e^2)*x/(b^2*c))/sqrt(c*x^2 + b*x) - e^2*log(abs(-2*(sqrt(c)*x - sqrt(
c*x^2 + b*x))*sqrt(c) - b))/c^(3/2)